If AD and PM are medians of triangles ABC and PQR,

Solution:

$\Delta \mathrm{ABC} \sim \Delta \mathrm{PQR}$ (Given)

$\Rightarrow \frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A C}{P R}$

$\angle \mathrm{A}=\angle \mathrm{P}, \angle \mathrm{B}=\angle \mathrm{Q}, \angle \mathrm{C}=\angle \mathrm{R}$ ...(1)

Now, $\quad B D=C D=\frac{1}{2} B C$

and $\quad Q M=R M=\frac{1}{2} Q R$ ...(2)

( $\because \mathrm{D}$ is mid-point of $\mathrm{BC}$ and $\mathrm{M}$ is mid-point of $\mathrm{QR}$ )



From (1), $\frac{A B}{P Q}=\frac{B C}{Q R} \Rightarrow \frac{A B}{P Q}=\frac{2 B D}{2 Q M}(B y(2))$

$\Rightarrow \frac{A B}{P Q}=\frac{B D}{Q M}$

Thus, we have $\frac{A B}{P Q}=\frac{B D}{O M}$

and $\angle \mathrm{ABD}=\angle \mathrm{PQM} \quad(\because \angle \mathrm{B}=\angle \mathrm{Q})$

$\Rightarrow \Delta \mathrm{ABD} \sim \Delta \mathrm{PQM} \quad$ (By SAS similarity criterion)

$\Rightarrow \frac{A B}{P Q}=\frac{A D}{P M}$